Divisibility Puzzle

Date: 01/22/2001 at 23:38:51
From: Li
Subject: Math contest

Dr. Math: 

I am a high school student in grade 10 who is in an enriched math 
class. Soon I will enter a math contest. Before I ask any specific 
questions, could you please give me some suggestions about how to get 
myself prepared?

Here is my question: The leftmost digit of an integer of length digits 
is 3. In this integer, any two consecutive digits must be divisible by 
17 or 23. The 2000th digit may be either "a" or "b" . What is the 
value of a+b? 

In this question, does consecutive just mean adjacent numbers? If the 
number starts with 3, then it will be impossible for any consecutive 
number to be divisible by 17 or 23. 

I really hope you can point me in a direction to improve my math 
solving skills. Are fundamentals more important than techniques? 

Yours truly, 
A sincere visitor to your Web site

Date: 01/23/2001 at 16:20:36
From: Doctor Greenie
Subject: Re: Math contest 

Hello, Li -

Certainly you can't be successful in mathematics, whether in contests 
or in application to a regular job, if you don't have a good, sound 
understanding of the fundamentals of mathematics. Although we can 
program computers to perform the calculations related to most math 
fundamentals, we can't yet program them to think analytically and 
devise techniques for solving problems, so in my opinion, techniques 
and problem-solving skills are more important than fundamentals.

As for preparation for math contests specifically, my primary 
suggestion would be to be confident. Be proud of the math skills you 
have, and don't worry about the fact that there are some math skills 
you don't have yet. And remember that this is (presumably) only a 
friendly competition. If you don't place well in the competition, you 
will at least have gained some valuable experience from having 
competed.

And one really nice thing about math competitions as I remember them 
from my experience at your age is that you have a chance to meet new 
friends who share your interest and ability in math.

Now, on to your question...

I'm sure that what they mean is that the 2-digit number formed by any 
two adjacent digits in the number is divisible by either 17 or 23.

The 2-digit numbers that are divisible by either 17 or 23 are the 
following:

    17, 23, 34, 46, 51, 68, 69, 85, and 92

Note that there are two 2-digit numbers in this list with first digit 
6 and none with first digit 7; for any other first digit there is only 
one entry in the list. (We don't need to worry about first digit 0, 
because no 2-digit numbers that are multiples of 17 or 23 have last 
digit 0.)  This list of numbers means that the next digit is always 
uniquely determined if the current digit is anything other than 0, 6, 
or 7. It also means that if the digit 7 appears in the number, then it 
is the last digit of the number; and it means that whenever the 
current digit is 6, there are two possibilities for the next digit.

Since the first digit is 3, then according to the list of numbers 
above, the next digit must be 4, to make the 2-digit number 34.  Then 
the third digit must be 6, in order to make the 2-digit number in the 
list with first digit 4.

Now, with first digit 6 in the next 2-digit integer, there are two 
possibilities for the next digit: 8 or 9. Continuing developing each 
of these integers in this manner, we get the following two 
possibilities for the integer:

    (1) 3468517....
    (2) 346923469234692....

The first of these strings leads to a dead end, because there is no 
2-digit number divisible by either 17 or 23 with first digit 7.

The second possibility gives us an integer consisting of a sequence of 
the 5 repeating digits 34692.

So all the integers that satisfy the conditions of the problem will be 
one of two types:

  (i) 3469234692....  with these digits repeating to the last digit; 

or

 (ii) 34692...3468517  with the repeating digits repeating any number
            of times but with the number ending in one or more of the
            digits "8517" following the final 6

So for most integers that satisfy the conditions of the problem, the 
2000th digit is 2; and for a very few such integers (which end shortly 
after the 2000th digit) the 2000th digit is 5.

The final answer to the question that was asked is, then, 2 + 5 = 7.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/